Differentiate the following function w.r.t. x:
y = ${(sinx)}^{x} + \sin^{- 1} \sqrt{x}$

Solution

$Y = {(\sin x)}^{x} + \sin^{- 1} \sqrt{x} Let u = {(\sin x)}^{x} and v = \sin^{- 1} \sqrt{x}$

Now y = u + v

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} . . . . . . (i) Consider u = {(\sin x)}^{x}$

Taking logarithms on both sides, we have,

log u = x log ( sin x )

Differentiating with respect to x, we have,

$\frac{1}{u} . \frac{du}{dx} = \log (\sin x) + \frac{x}{\sin x} . \cos x \Rightarrow \frac{du}{dx} = {(\sin x)}^{x} (\log (\sin x) + x \cot x) . . . . (ii) Consider v = \sin^{- 1} \sqrt{x} \frac{dv}{dx} = \frac{1}{\sqrt{1 - x}} x \frac{1}{2 \sqrt{x}} . . . . (iii)$

From (i), (ii) and (iii)

We get, $\frac{dy}{dx} = {(\sin x)}^{x} (\log (\sin x) + x \cot x) + \frac{1}{2 \sqrt{x} . \sqrt{1 - x}}$

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Differentiate the following function w.r.t. x:
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